# Linear First Order ODE/(2 y - x^3) dx = x dy

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## Theorem

$(1): \quad \paren {2 y - x^3} \rd x = x \rd y$

has the general solution:

$y = -x^3 + C x^2$

## Proof

Rearranging $(1)$:

$(2): \quad \dfrac {\d y} {\d x} - \dfrac 2 x y = - x^2$

$(2)$ is in the form:

$\dfrac {\d y} {\d x} + \map P x y = \map Q x$

where $\map P x = -\dfrac 2 x$.

Thus:

 $\ds \int \map P x \rd x$ $=$ $\ds \int -\dfrac 2 x \rd x$ $\ds$ $=$ $\ds -2 \ln x$ $\ds$ $=$ $\ds \map \ln {\frac 1 {x^2} }$ $\ds \leadsto \ \$ $\ds e^{\int P \rd x}$ $=$ $\ds \frac 1 {x^2}$

Thus from Solution by Integrating Factor, $(1)$ can be rewritten as:

$\map {\dfrac \d {\d x} } {\dfrac y {x^2} } = -1$

and the general solution is:

$\dfrac y {x^2} = -x + C$

or:

$y = -x^3 + C x^2$

$\blacksquare$